Finding Critical Numbers Of F(x) = X^(4/5)(x-6)^2

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Introduction to Critical Numbers

In calculus, critical numbers, also known as critical values or critical points, play a pivotal role in understanding the behavior of a function. Specifically, they help us identify potential local maxima, local minima, and saddle points. A critical number of a function f is a value c in the domain of f such that either f'(c) = 0 or f'(c) does not exist. These points are crucial for analyzing the function's increasing and decreasing intervals, concavity, and overall shape. By locating critical numbers, we can gain deeper insights into the function's characteristics and its graphical representation. The process of finding critical numbers typically involves computing the derivative of the function, setting it equal to zero, and solving for the variable. Additionally, we must consider points where the derivative is undefined, as these points may also be critical. The information gleaned from critical numbers is fundamental in various applications, such as optimization problems in physics, engineering, and economics, where finding maximum or minimum values is essential. Understanding critical numbers is a cornerstone of differential calculus and is essential for students and professionals alike in various fields.

Problem Statement

Our task is to find the critical numbers of the function $F(x) = x{\frac{4}{5}}(x-6)2$. This involves several steps, including finding the derivative of the function, setting the derivative equal to zero, and identifying any points where the derivative is undefined. Critical numbers are essential for understanding the behavior of a function, as they indicate potential local maxima, local minima, or inflection points. Therefore, accurately determining these numbers is crucial for further analysis and applications of the function. The given function is a product of a power function and a polynomial function, which means we'll need to apply the product rule and chain rule when finding the derivative. The power function part, $x^{\frac{4}{5}}$, introduces the possibility of points where the derivative might be undefined, particularly at x = 0, due to the fractional exponent. The polynomial part, $(x-6)^2$, will contribute to the critical numbers where the derivative equals zero. By systematically working through the differentiation process and carefully analyzing the resulting equation, we can identify all critical numbers of the given function and lay the groundwork for further analysis of its properties. This exercise is a fundamental application of differential calculus and highlights the importance of mastering differentiation techniques for understanding function behavior.

Step-by-Step Solution

1. Find the Derivative of F(x)

To begin, we need to find the derivative of the function $F(x) = x{\frac{4}{5}}(x-6)2$. This requires the application of the product rule, which states that if we have a function $F(x) = u(x)v(x)$, then the derivative $F'(x) = u'(x)v(x) + u(x)v'(x)$. In our case, we can identify $u(x) = x^{\frac{4}{5}}$ and $v(x) = (x-6)^2$. Next, we find the derivatives of $u(x)$ and $v(x)$. For $u'(x)$, we use the power rule, which states that if $u(x) = x^n$, then $u'(x) = nx^{n-1}$. Applying this, we get:

uβ€²(x)=45x45βˆ’1=45xβˆ’15u'(x) = \frac{4}{5}x^{\frac{4}{5}-1} = \frac{4}{5}x^{-\frac{1}{5}}

For $v'(x)$, we use the chain rule. If $v(x) = (x-6)^2$, then $v'(x) = 2(x-6) \cdot 1 = 2(x-6)$. Now, we can apply the product rule to find $F'(x)$:

Fβ€²(x)=uβ€²(x)v(x)+u(x)vβ€²(x)=45xβˆ’15(xβˆ’6)2+x45β‹…2(xβˆ’6)F'(x) = u'(x)v(x) + u(x)v'(x) = \frac{4}{5}x^{-\frac{1}{5}}(x-6)^2 + x^{\frac{4}{5}} \cdot 2(x-6)

2. Simplify the Derivative

To make the derivative easier to work with, we need to simplify it. Let's factor out common terms. We can factor out $x^{-\frac{1}{5}}$ and $(x-6)$:

Fβ€²(x)=xβˆ’15(xβˆ’6)[45(xβˆ’6)+2x]F'(x) = x^{-\frac{1}{5}}(x-6) \left[ \frac{4}{5}(x-6) + 2x \right]

Now, we simplify the expression inside the brackets:

45(xβˆ’6)+2x=45xβˆ’245+2x=45x+105xβˆ’245=145xβˆ’245\frac{4}{5}(x-6) + 2x = \frac{4}{5}x - \frac{24}{5} + 2x = \frac{4}{5}x + \frac{10}{5}x - \frac{24}{5} = \frac{14}{5}x - \frac{24}{5}

So, the simplified derivative is:

Fβ€²(x)=xβˆ’15(xβˆ’6)(145xβˆ’245)F'(x) = x^{-\frac{1}{5}}(x-6) \left(\frac{14}{5}x - \frac{24}{5} \right)

We can further simplify by factoring out $\frac{2}{5}$ from the term inside the parentheses:

Fβ€²(x)=25xβˆ’15(xβˆ’6)(7xβˆ’12)F'(x) = \frac{2}{5}x^{-\frac{1}{5}}(x-6)(7x - 12)

3. Find Where F'(x) = 0

To find the critical numbers, we need to determine where the derivative $F'(x)$ is equal to zero. Setting $F'(x) = 0$, we have:

25xβˆ’15(xβˆ’6)(7xβˆ’12)=0\frac{2}{5}x^{-\frac{1}{5}}(x-6)(7x - 12) = 0

This equation is satisfied if any of the factors are zero. Thus, we have three cases to consider:

  1. x^{-\frac{1}{5}} = 0$: This term is never zero because $x^{-\frac{1}{5}} = \frac{1}{x^{\frac{1}{5}}}$, and a fraction can only be zero if its numerator is zero.

  2. x - 6 = 0$: This gives us $x = 6

  3. 7x - 12 = 0$: Solving for $x$, we get $7x = 12$ or $x = \frac{12}{7}

So, the values of $x$ that make the derivative zero are $x = 6$ and $x = \frac{12}{7}$.

4. Find Where F'(x) is Undefined

We also need to find the values of $x$ for which the derivative $F'(x)$ is undefined. The derivative is given by:

Fβ€²(x)=25xβˆ’15(xβˆ’6)(7xβˆ’12)F'(x) = \frac{2}{5}x^{-\frac{1}{5}}(x-6)(7x - 12)

This derivative is undefined when $x^{-\frac{1}{5}}$ is undefined, which occurs when $x = 0$. The term $x^{-\frac{1}{5}}$ can be written as $\frac{1}{\sqrt[5]{x}}$, and this is undefined when the denominator is zero, i.e., when $x = 0$. Therefore, $x = 0$ is another critical number.

5. List the Critical Numbers

Combining the values where $F'(x) = 0$ and where $F'(x)$ is undefined, we find the critical numbers to be:

x=0,127,6x = 0, \frac{12}{7}, 6

These are the points where the function may have local maxima, local minima, or saddle points. To fully understand the behavior of the function, further analysis, such as using the first derivative test or the second derivative test, would be necessary.

Conclusion

In conclusion, by systematically applying the product rule and chain rule to find the derivative of the function $F(x) = x{\frac{4}{5}}(x-6)2$, we identified the critical numbers. We found that the derivative is zero at $x = 6$ and $x = \frac{12}{7}$, and it is undefined at $x = 0$. Therefore, the critical numbers of the function are $0, \frac{12}{7},$ and $6$. These critical numbers are crucial for further analysis of the function's behavior, such as determining intervals of increasing and decreasing, local maxima and minima, and concavity. The process of finding critical numbers is a fundamental technique in calculus, and this problem provides a clear example of its application. Understanding and accurately finding critical numbers is essential for solving various problems in mathematics, physics, engineering, and other fields where optimization and analysis of functions are required. This comprehensive step-by-step solution demonstrates the methodical approach needed to tackle such problems, emphasizing the importance of both differentiation techniques and careful algebraic manipulation.

Final Answer: The final answer is 0,12/7,6\boxed{0, 12/7, 6}